How to solve this linear ODE? Can somebody show me how to solve the following ODE:
$$
t^2 y''+y'=0.
$$
I would be thankful for any help
 A: The hint was given by Robert Z. in his answer.
$$t^2 z'+z=0 \implies z=c_1 e^{\frac{1}{t}}$$ So $$y=c_1\int  e^{\frac{1}{t}}\,dt$$ Integrating by parts $$u=e^{\frac{1}{t}}\implies u'=-\frac{e^{\frac{1}{t}}}{t^2}$$ $$\int  e^{\frac{1}{t}}\,dt=e^{\frac{1}{t}}+\int\frac{e^{\frac{1}{t}}}{t}\,dt$$ For the last integral, set $t=\frac 1x$ to get $$\int\frac{e^{\frac{1}{t}}}{t}\,dt=-\int \frac{e^x}{x}\,dx=-\text{Ei}(x)$$ where appears the exponential integral.  So, finally,$$y=c_1 \left(t\,e^{\frac{1}{t}} -\text{Ei}\left(\frac{1}{t}\right)\right)+c_2$$
A: Hint. Let $z(t)=y'(t)$ and separate variables.
A: Without substitution: You can start dividing by $t$ for $t \neq 0,$ which yields
$$y''+\frac{1}{t^2}y'=0.$$
Now the Integrating factor is $I.F.=e^{\int{1/{t^2}}~dt}=e^{-1/{t}},$ which gives
$$\frac{d}{dt}\left( e^{-1/{t}} \cdot y'  \right)=0,$$ which yields
$$y'(t)=C \cdot e^{1/t}.$$
Now, using integration by parts yields the solution above with two arbitrary constants. 
Further, for $t=0,$ we simply get $y=\textrm{constant}.$
